\(\frac{3}{4};\frac{10}{3};\frac{9}{5};\frac{18}{2};\frac{71}{100}\)hãy viết dưới dạng thương
TX
Những câu hỏi liên quan
\(\frac{1}{\frac{2}{\frac{3}{\frac{4}{...}+\frac{4}{...}}+\frac{3}{\frac{4}{...}+\frac{4}{...}}}+\frac{2}{\frac{3}{\frac{4}{...}+\frac{4}{...}}+\frac{3}{\frac{4}{...}+\frac{4}{...}}}}\)
Bài 1:Tính
\(\frac{\frac{1}{3}-\frac{4}{5}}{\frac{1}{3}+\frac{4}{5}}.\frac{\frac{3}{4}-\frac{5}{3}}{\frac{3}{4}+\frac{5}{3}}:\frac{\frac{4}{5}-1}{1-\frac{2}{3}}\)
Bài làm ai trên 11 điểm tích mình thì mình tích lại
Ông tùng hơn tùng số tuổi là :
29 + 32 = 61 (tuổi )
Vậy ông của tùng hơn tùng 61 tuổi
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1,\(\frac{4}{3}-\frac{7}{12}+\left(\frac{3}{8}-4x+\frac{1}{2}\right)=|\frac{4}{3}+|-\frac{1}{2}||-\frac{3}{4}\)
2,\(|2x-\frac{1}{3}|+\frac{4}{5}-\frac{3}{2}+\frac{7}{3}=|\frac{4}{5}-\frac{5}{3}|+\frac{1}{4}\)
B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
t tưởng mọi hôm bài này m làm thạo lắm mà bây h chịu ak
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B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=-1\frac{1}{5}\cdot\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}\div\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=\frac{-6}{5}\cdot4\div\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(B=\frac{-24}{5}\div\frac{4}{5}\)
\(B=-6\)
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\(B=-1\frac{1}{5}.\frac{4.\frac{3}{7}}{\frac{3}{37}}:\frac{4+3.\frac{4}{1}}{5+3.\frac{5}{1}}\)
\(B=-\frac{6}{5}.\frac{148}{7}:\frac{4}{5}\)
\(B=-\frac{222}{7}\)
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Bài 1:Tính
\(\frac{\frac{1}{3}-\frac{4}{5}}{\frac{1}{3}+\frac{4}{5}}.\frac{\frac{3}{4}-\frac{5}{3}}{\frac{3}{4}+\frac{5}{3}}:\frac{\frac{4}{5}-1}{1-\frac{2}{3}}\)
\(\dfrac{\dfrac{1}{3}-\dfrac{4}{5}}{\dfrac{1}{3}+\dfrac{4}{5}}.\dfrac{\dfrac{3}{4}-\dfrac{5}{3}}{\dfrac{3}{4}+\dfrac{5}{3}}:\dfrac{\dfrac{4}{5}-1}{1-\dfrac{2}{3}}\)
\(=\dfrac{\dfrac{-7}{15}}{\dfrac{17}{15}}.\dfrac{-\dfrac{11}{12}}{\dfrac{29}{12}}:\dfrac{\dfrac{-1}{5}}{\dfrac{1}{3}}\)
\(=\dfrac{-7}{17}.\dfrac{-11}{29}:\left(-\dfrac{3}{5}\right)\)
\(=\dfrac{77}{493}:\left(-\dfrac{3}{5}\right)\)
\(=-\dfrac{385}{1479}\)
Vậy ...
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\(\frac{\frac{-2}{3}+\frac{3}{4}-2}{\frac{2}{3}-\frac{3}{4}+2}-\frac{\frac{-2}{3}-\frac{3}{4}-2}{\frac{2}{3}+\frac{2}{4}+2}\)
\(=\frac{-1\left(\frac{2}{3}-\frac{3}{4}+2\right)}{\frac{2}{3}-\frac{3}{4}+2}-\frac{-1\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=-1-\left(-1\right)\)
\(=-1+1\)
\(=0\)
\(=\frac{-\left(\frac{2}{3}+\frac{3}{4}-2\right)}{\frac{2}{3}+\frac{3}{4}-2}-\frac{-\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\left(-1\right)-\left(-1\right)\)
\(=0\)
\(-1\frac{1}{5}.\frac{4.\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
Tính
\(\frac{-\frac{2}{3}+\frac{3}{4}-2}{\frac{2}{3}-\frac{3}{4}-2}\cdot\frac{-\frac{2}{3}-\frac{3}{4}-2}{\frac{2}{3}+\frac{3}{4}+2}\)
Ta có \(\frac{-\frac{2}{3}+\frac{3}{4}-2}{\frac{2}{3}-\frac{3}{4}-2}.\frac{-\frac{2}{3}-\frac{3}{4}-2}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\frac{-\left(\frac{2}{3}-\frac{3}{4}+2\right)}{\frac{2}{3}-\frac{3}{4}-2}.\frac{-\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\frac{23}{25}.\left(-1\right)\)
\(=\frac{-23}{25}\)
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Thực hiên phép tính:
\(\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
Sửa đề; \(\dfrac{1}{5}\cdot\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{1}{5}\cdot4:\dfrac{4}{5}=\dfrac{4}{5}\cdot\dfrac{5}{4}=1\)
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